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In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.
Prove that $ar(\triangle LOB) = ar(\triangle MOC).
Given:
In a $\triangle ABC$, $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.
To do:
We have to prove that $ar(\triangle LOB) = ar(\triangle MOC)$.
Solution:
Join $LM, LC$ and $MB$.
$L$ and $M$ are the mid points of $AB$ and $AC$.
This implies,
$LM \| BC$
$\triangle LBM$ and $\triangle LCM$ are on the same base $LM$ and between the same parallels.
Therefore,
$ar(\triangle LBM) = ar(\triangle LCM)$......…(i)
$ar(\triangle LCM) = ar(\triangle LBM)$
$\triangle LBC$ and $\triangle MBC$ are on the same base $BC$ and between the same parallels.
Therefore,
$ar(\triangle LBC) = ar(\triangle MBC)$......…(ii)
$ar(\triangle LMB) = ar(\triangle LMC)$ [From (i)]
$ar(\triangle ALM) + ar(\triangle LMB) = ar(\triangle ALM) + ar(\triangle LMC)$ [Adding $ar(\triangle ALM)$ on both sides]
$ar(\triangle ABM) = ar(\triangle ACL)$
$ar(\triangle LBC) = ar(\triangle MBC)$ [From (ii)]
$ar(\triangle LBC) - ar(\triangle BOC) = ar(\triangle MBC) - ar(\triangle BOC)$
$ar(\triangle LBO) = ar(\triangle MOC)$
Hence proved.
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